A body initially at rest, starts moving with a constant acceleration of 2ms-2 .calculate the velocity acquired and the distance travelled in 5s

The velocity acquired in 5 s is 10 m/s and the distance travelled in 5 s is 25 m.

Explanation:

As acceleration is the measure of change in velocity with respect to time, if the acceleration and time is known , we can easily determine the velocity using the first equation of motion.

v = u +at

Here v is the final velocity which we have to determine, u is the initial velocity which is given as zero and acceleration is given as 2 ms⁻².

So, velocity of the body at 5 s will be v = 0 + (2× 5 s), v = 10 m/s.

As the given problem stated that motion is occurring at constant acceleration, then the distance travelled by the body can be obtained from second equation of motion.

s = ut + 0.5 at²

So, here the displacement s will be equal to the distance as the starting position is considered as zero. And initial velocity u is also zero, a is given as 2 ms⁻² and the time is 5s.

Then, s = (0×5)+(0.5×2×5×5)=25 m

So, the velocity acquired in 5 s is 10 m/s and the distance travelled in 5 s is 25 m.

air exerts forces on falling objects near earth’s surface.

explanation:

the term 'free falls' refer to object that are falling towards the earth, in a situation in which gravity is the only force acting on the object.

for an object in free fall, the acceleration is constant (g, the gravitational acceleration), and so the velocity of the object constantly increases.

for objects near earth's surface, there is always another force acting on the object: the resistance due to the air. this force is opposed to the direction of fall of the object, and it is proportional to the velocity of the object, so as the object falls down, this force increases up to a point when it balances the force of gravity, and the object continues its motion at constant velocity (called terminal velocity). since in this case gravity is not the only force acting on the object, we are not in a situation of free fall.

distance = 150m

displ = root (50 ^2 + 100^2)

root(2500+10000)

root 125000